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Alla [95]
3 years ago
14

A circle is translated 4 units to the right and then reflected over the x axis. What statement will always be true of the circle

at the new location?
Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

The statements are not given, so i will answer it in the most general way possible:

First, let's recall two things:

Horizontal translation.

For a number A positive, we can move the graph of f(x) by A units to the right by the transformation g(x) = f(x - A)

Reflection over x

For a point (x, y), a reflection over the x-axis transforms this point into (x, - y).

Now let's analyze the problem:

A generic circle, of radius R and centered in the point (a, b) can be written as:

(x - a)^2 + (y - b)^2 = R^2.

Now let's translate this 4 units to the right:

(x - 4 - a)^2 + (y - b)^2 = R^2.

or in standard circle notation:

(x - (4 + a))^2 + (y - b)^2 = R^2.

Now the circle is centered in the point (a + 4, b).

Now let's do a reflection over the x-axis.

This means that the sign of y changes, so now we have:

(x - (4 + a))^2 + (-y - b)^2 = R^2.

But because the term with y is squared, this is the same than:

(x - (4 + a))^2 + (y + b)^2 = R^2.

Or, in standard circle notation:

(x - (4 + a))^2 + (y - (-b))^2 = R^2.

Then the circle is now centered in point (a + 4, -b).

So now you know where the circle is located, and also you can see that the radius of the circle never changed.

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Step-by-step explanation:

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Let’s split that into two sections so it’s easier to see:

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Next let’s look at what 15x^3 and -25x have in common. They have 5x in common.

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Next let’s look at what -6x^2and 10 have in common. They only have 2 in common, so we factor out 2.

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So the end result will be : 5x(3^2-5)-2(3^2-5)

And the complete factorization will be (5x-2)(3^2-5)

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